Surface Chemistry - Result Question 8

8. Adsorption of a gas follows Freundlich adsorption isotherm. $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. The plot of $\log \frac{x}{m}$ versus $\log p$ is shown in the given graph. $\frac{x}{m}$ is proportional to

(2019 Main, 8 April I)

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-199_jpg_height_278_width_328_top_left_y_1597_top_left_x_1423.jpg"width="250">

(a) $p^{2 / 3}$

(b) $p^{3 / 2}$

(c) $p^{3}$

(d) $p^{2}$

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Answer:

Correct Answer: 8. (d)

Solution:

$ \begin{aligned} & \text { Key Idea According to Freundlich, } \\ & \qquad \frac{x}{m}=K p^{1 / n}[n>1] \end{aligned} $

where, $m=$ mass of adsorbent, $x=$ mass of the gas adsorbed, $\frac{x}{m}=$ amount of gas adsorbed per unit mass of solid adsorbent, $p=$ pressure, $K$ and $n=$ constants.

The logarithm equation of Freundlich adsorption isotherm is

$ \log \frac{x}{m}=\log K+\frac{1}{n} \log p $

On comparing the above equation with straight line equation, $(y=m x+c)$

we get

$ m=\text { slope }=\frac{1}{n} \quad \text { and } \quad c=\log K $

From the given plot,

$m=\frac{y _2-y _1}{x _2-x _1} =\frac{1}{n}=\frac{2}{3} $

$\therefore \frac{x}{m} =K p^{2 / 3}$



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